Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
// I - Coins
#include <bits/stdc++.h>
using namespace std;
using dl = double;

int main(){
	int N; cin>>N;

	vector<dl> P(N+1); for(int i=1; i<=N; i++) cin>>P[i];

	int C = (N+1)/2;//ceil
	vector<dl> dp(N+2);
	dp[1] = 1.;
	for(int lv=1; lv<=N; lv++){
		dl f = P[lv], b = 1.-f;
		int e = max(1, lv-C+2);
		for(int i=lv+1; i>=e; i--) dp[i] = b*dp[i] + f*dp[i-1];
	}

	dl a = 0; for(int i=N+1; i>C; i--) a += dp[i];
	printf("%.10g\n",a);
}